A block of mass 1.5 kg moves along the floor of a hall with a speed of 5 ms⁻¹ . It strikes an uncompressed spring and compresses it till the block becomes motionless. If the force constant of the spring is 10000 Nm ⁻¹ and the spring is compressed by 5cm, calculate the effective force of kinetic friction.

Step-by-step solution

Initial kinetic energy of the block is K_i = 1 2 mv^2 . Final kinetic energy of the block is K_f = 0 . Work done by the spring force is W_s = - 1 2 kx^2 . Work done by the kinetic friction is W_f = -f_k x . Using the work-energy theorem, W_s + W_f = K_f - K_i . - 1 2 kx^2 - f_k x = 0 - 1 2 mv^2 f_k x = 1 2 mv^2 - 1 2 kx^2 Substituting the given values: m = 1.5 kg, v = 5 m/s, k = 10000 N/m, x = 0.05 m. f_k (0.05) = 1 2 (1.5)(5)^2 - 1 2 (10000)(0.05)^2 f_k (0.05) = 18.75 - 12.5 f_k (0.05) = 6.25 f_k = 6.25 0.05 = 125